a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy...
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(1-3x\right)=0\)
\(\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy..
d) \(3x\left(x-1\right)+\left(1-x\right)^2=0\)
\(\Rightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(3x-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy..
c) \(2\left(x-4\right)^2-\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\cdot\left[2\left(x-4\right)-1\right]=0\)
\(\Leftrightarrow\left(x-4\right)-\left(2x-8-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy \(x_1=4;x_2=\dfrac{9}{2}\)
d) \(3x\left(x-1\right)+\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(3+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2+x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2+x=0\end{matrix}\right.\)
Vậy \(x_1=-2;x_2=1\)
d) \(5x\left(x-9\right)^2-\left(9-x\right)^3=0\)
\(\Rightarrow5x\left(x^2-18x+81\right)-\left(729-243x+27x^2-x^3\right)=0\)
\(\Leftrightarrow5x^3-90x^2+405x-729+243x-27x^2+x^3=0\)
\(\Leftrightarrow6x^3-117x^2+648x-729=0\)
\(\Leftrightarrow3\left(2x^3-39x^2+216x-243\right)=0\)
\(\Leftrightarrow3\left(2x^3-18x^2-21x^2+189x+27x-243\right)=0\)
\(\Leftrightarrow3\left(2x^2\cdot\left(x-9\right)-21\cdot\left(x-9\right)+27\left(x-9\right)\right)=0\)
\(\Leftrightarrow3\left(x-9\right)\left(2x^2-21x+27\right)=0\)
\(\Leftrightarrow3\left(x-9\right)\cdot\left(2x^2-38-18x+27\right)=0\)
\(\Leftrightarrow3\left(x-9\right)\cdot\left(x\cdot\left(2x-3\right)-9\left(2x-3\right)\right)=0\)
\(\Leftrightarrow3\left(x-9\right)\cdot\left(x-9\right)\cdot\left(2x-3\right)=0\)
\(\Leftrightarrow3\left(x-9\right)^2\cdot\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-9\right)^2\cdot\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-9\right)^2=0\\2x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{3}{2};x_2=9\)
b) \(2x\left(x-4\right)^2-\left(x-4\right)=0\)
\(\Rightarrow\)\(\left(x-4\right)\left(2x^2-8x-1\right)=0\)
( Vì \(2x^2-8x-1>0\) )
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
Vậy..
( câu này hơi không chắc :v )
e) \(5x\left(x-9\right)^2-\left(9-x\right)^3=0\)
\(\Rightarrow\)\(5x\left(x-9\right)^2+\left(x-9\right)^3=0\)
\(\Rightarrow\left(x-9\right)^2\left(5x+x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-9\right)^2=0\\6x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
a, \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b, \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(1-3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
Các câu còn lại thì bn khác làm r nhé!